1 (53)

Prévia do material em texto

Principles of Instrumental Analysis, 6th ed. Chapter 9
 
 5
9-14. The energies of the 3p states can be obtained from the emission wavelengths shown in 
Figure 8-1. For Na, we will use an average wavelength of 5893 Å and for Mg+, 2800 Å. 
 For Na, the energy of the excited state is 
 
34 8 1
19
3 , 10
6.626 10 J s 3.00 10 m s = = 3.37 10 J
5893 Å 10 m/Åp Na
hcE
λ
− −
−
−
× × ×
= ×
×
 
 For Mg+ 
 +
34 8 1
19
103 ,Mg
6.626 10 J s 3.00 10 m s= = 7.10 10 J
2800 Å 10 m/Åp
E
− −
−
−
× × ×
×
×
 
 (a) Substituting into Equation 8-1, gives at 2100 K 
 
19
5
23 1
0 Na
3.37 10 J = 3 exp = 2.67 10
1.38 10 J K 2100K
jN
N
−
−
− −
⎛ ⎞ ⎛ ⎞×
− ×⎜ ⎟ ⎜ ⎟× ×⎝ ⎠⎝ ⎠
 
 
+
19
11
23 1
0 Mg
7.10 10 J = 3 exp = 6.87 10
1.38 10 J K 2100 K
jN
N
−
−
− −
⎛ ⎞ ⎛ ⎞×
− ×⎜ ⎟ ⎜ ⎟× ×⎝ ⎠⎝ ⎠
 
 Proceeding in the same wave we obtain for Na and Mg+ 
 (b) Nj/N0 = 6.6 × 10–4 and 5.9 × 10–8 
 (c) Nj/N0 = 0.051 and 5.7 × 10–4 
9-15. The energy difference between the 3p and 3s states was shown in Solution 8-9 to be E = 
3.37 × 10-19 J. The energy difference between the 4s and 3p states E´ can be calculated 
from the wavelength of the emission lines at 1139 nm 
 
34 8 1
19
9
6.626 10 J s 3.00 10 m s = = 1.75 10 J
1139 nm 10 m/nm
E
− −
−
−
× × ×′ ×
×
 
 The energy difference between the 4s and 3s state is then 
 E′′ = 3.37 × 10–19 + 1.75 × 10–19 = 5.12 × 10–19 J